Max and Min of $f(x,y)$ Let $f(x,y)=x(y \log y-y)-y \log x$. Find $\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})$. 
 A: EDIT: I have a feeling I'm missing some technique, is there another method I seem unaware of? /EDIT (By the way, from the graph, I'd guess it is at (x,y)=(.5,.5))
First, find the function of $x$, 
$$m(x)=\min_{y\in[.5,1]}f(x,y).$$
Fix $x$ and find $m(x)$ by any means necessary. We could use calculus to find the critical points, but the algebra looks a little tedious. Let's use a graphical program to get a clue: 
http://www.wolframalpha.com/input/?i=plot%28x%28ylogy%E2%88%92y%29%E2%88%92ylogx%2C+.5%3Cx%3C2%2C+.5%3Cy%3C1%29
From the linked plot, if we move in the $y$-direction, at $x=.5$, it is an increasing function, while at $x=2$, it is a decreasing function. Let's just check for critical points after all. We set
$$f_y(x,y)=x\log y-\log x=0,$$ 
so for any $x$, critical points along $y$ occur at they $y$ value $y=x^{1\over x}$. To find the minimum (at fixed $x$), we compare $f(x,x^{1\over x})$, $f(x,.5)$ and $f(x,1)$. 

A: Note that $$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)}) \leq \smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}f(x,\frac{1}{2})=\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})$$
Since $-\frac{x\log 2+x+\log x}{2}$ is a decreasing function of $x$, we have 
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(-\frac{x\log 2+x+\log x}{2})=-\frac{\frac{1}{2}\log 2+\frac{1}{2}+\log \frac{1}{2}}{2}=\frac{\log 2-1}{4}$$
with equality when $x=\frac{1}{2}$. We claim that equality holds throughout and thus
$$\smash{\displaystyle\max_{\frac{1}{2} \leq x \leq 2}}(\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(x,y)})=\frac{\log 2-1}{4}$$
It now suffices to show that 
$$\smash{\displaystyle\min_{\frac{1}{2} \leq y \leq 1} f(\frac{1}{2},y)}=f(\frac{1}{2}, \frac{1}{2})$$
We have 
$$f(\frac{1}{2}, y)=\frac{1}{2}y\log y-\frac{1}{2}y+y\log 2$$
$$\frac{d f(\frac{1}{2}, y)}{dy}=\frac{1}{2}\log y+\log 2>\frac{1}{2}\log \frac{1}{2}+\log 2=\frac{1}{2}\log 2>0$$
Therefore $f(\frac{1}{2}, y)$ is a non-decreasing function of $y$, and we are done.
A: For fixed $\frac12\le x\le 2,$ the function $g_x(y)=f(x,y)$ will be a differentiable function in the variable $y$ on the positive reals. Then $g_x(y)$ can only achieve its minimum in the closed interval $\left[\frac12,2\right]$ at the boundary, or at a critical point in its interior.
Now, for $y>0$ we have $$g_x'(y)=\frac{\partial}{\partial y}\left[f(x,y)\right]=x\frac{\partial}{\partial y}\left[y\log y\right]-x-\log x=x\log y-\log x=\log\frac{y^x}x,$$ which has its sole zero when $y^x=x$. Since $$g''_x(y)=\frac xy$$ is positive for all $y>0,$ then $g_x$ has a global minimum when $y^x=x$ and $y>0,$ namely $$\begin{align}g_x(x^{1/x}) &= x\left(x^{1/x}\log(x^{1/x})-x^{1/x}\right)-x^{1/x}\log x\\ &= x\left(x^{1/x}\frac1x\log x-x^{1/x}\right)-x^{1/x}\log x\\ &= -x^{1+\frac1x}.\end{align}$$ You should verify that $x\mapsto-x^{1+\frac1x},$ $x\mapsto g_x(1/2),$ and $x\mapsto g_x(2)$ are decreasing functions on $[\frac12,2],$ so each will be maximized on that interval when $x=\frac12.$ Hence, $$\max_{\frac12\le x\le2}\min_{\frac12\le y\le2}g_x(y)=\min_{\frac12\le y\le2}g_{1/2}(y).$$ Note in particular that for $x=\frac12$, we have $x^{1/x}=\frac14,$ which is not in the interior of the closed interval. That is, $g_{1/2}(y)$ doesn't have a critical point in the interior of closed interval, so is minimized on the boundary, meaning $$\min_{\frac12\le y\le2}g_{1/2}(y)=\min\{g_{1/2}(1/2),g_{1/2}(2)\}=\min\{f(1/2,1/2),f(1/2,2)\}.$$ I'll let you take it from there.
